Question

Complete solution set of $ta{n}^2\left(si{n}^{-1}x\right)>1$ is :

• A. $(-1-\frac{1}{\sqrt{2}})\cup (\frac{1}{\sqrt{2}},1)$
• B. $(-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})-\left\{0\right\}$
• C. $(-1,1)-\left\{0\right\}$
• D. None of these

Answer 1

The correct option is A $(-1-\frac{1}{\sqrt{2}})\cup (\frac{1}{\sqrt{2}},1)$

$ta{n}^2\left(si{n}^{-1}x\right)>1$
$\Rightarrow tan\left(si{n}^{-1}x\right)<-1$ or $tan\left(si{n}^{-1}x\right)>1$
$\Rightarrow -\frac{\pi }{2}<si{n}^{-1}x<-\frac{\pi }{4}$ or $\frac{\pi }{4}<si{n}^{-1}x<\frac{\pi }{2}$
$\Rightarrow -1<x<\frac{-1}{\sqrt{2}}$ or $\frac{1}{\sqrt{2}}<x<1$
$\Rightarrow xϵ(-1-\frac{-1}{\sqrt{2}})\cup (\frac{1}{\sqrt{2}},1)$